Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $24$ years; the standard deviation is $3.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living between $31.8$ and $35.7$ years.
Solution: $24$ $20.1$ $27.9$ $16.2$ $31.8$ $12.3$ $35.7$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $24$ years. We know the standard deviation is $3.9$ years, so one standard deviation below the mean is $20.1$ years and one standard deviation above the mean is $27.9$ years. Two standard deviations below the mean is $16.2$ years and two standard deviations above the mean is $31.8$ years. Three standard deviations below the mean is $12.3$ years and three standard deviations above the mean is $35.7$ years. We are interested in the probability of a tiger living between $31.8$ and $35.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the tigers will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of tigers between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular tiger living between $31.8$ and $35.7$ years is $\color{orange}{2.35\%}$.